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How do I find the pH of 0.25M NH4Cl (while using Kb of NH3 to calculate the Ka of NH4+)?

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How do I find the pH of 0.25M NH4Cl (while using Kb of NH3 to calculate the Ka of NH4+)?

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  1. NH4+ --> NH3  & H +

    Ka = [NH3] [H+] / [NH4+]

    Ka = Kwater / Kb = 1e-14 / 1.8e-5 = 5.56e-10

    5.56e-10 = [NH3] [H+] / [NH4+]

    5.56e-10 = [x] [x] / [0.25]

    x = [H+] = 1.18e-5

    your answer is pH = 4.93

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